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Flow Through an Expansion Valve

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Flow Through an Expansion Valve


Refrigerant R717


enters an expansion valve (of cross-sectional area 0.011 m2) at 11 bar, 330 K and 25 m s-1,


and leaves at 2 bar.

This application calculates the temperature and velocity of the refrigerant as it exits the valve.


The First Law of Thermodynamics states

`#mscripts(mi("Q",fontstyle = "normal"),mi("sys",fontstyle = "normal"),none(),none(),mo("."),none(),none())` = `#mscripts(mi("E",fontstyle = "normal"),mi("sys",fontstyle = "normal"),none(),none(),mo("."),none(),none())`+`#mscripts(mi("W",fontstyle = "normal"),mi("s",fontstyle = "normal"),none(),none(),mo("."),none(),none())`+`#mscripts(mi("m",fontstyle = "normal"),mi("out",fontstyle = "normal"),none(),none(),mo("."),none(),none())`*(h__out+(1/2)*v__out^2+g*z__out)-`#mscripts(mi("m",fontstyle = "normal"),mo("in"),none(),none(),mo("."),none(),none())`h__in+(1/2)*v__in^2+g*z__in



`#mscripts(mi("Q",fontstyle = "normal"),mi("sys",fontstyle = "normal"),none(),none(),mo("."),none(),none())` is the heat generated by the system


`#mscripts(mi("E",fontstyle = "normal"),mi("sys",fontstyle = "normal"),none(),none(),mo("."),none(),none())`is the rate of change of stored energy within the system


W__s is the rate of work done by the system (except flow work)


`#mscripts(mi("m",fontstyle = "normal"),mo("in"),none(),none(),mo("."),none(),none())` and `#mscripts(mi("m",fontstyle = "normal"),mi("out",fontstyle = "normal"),none(),none(),mo("."),none(),none())`are the mass flowrates into and out of the system


h__in and h__out are the specific enthalpies of the fluid entering and leaving the system


v__in and v__out are the velocities of the fluid entering and leaving the system


z__in and z__out are the elevations of the fluid entering and leaving the system


For steady-state flow through an adiabatic expansion valve and no heat or work effects, the First Law of Thermodynamics reduces to


Assuming the input conditions are known, mass conversation implies that "(m)[in] =(m)[out] = m, "and hence

m = A*v__out*`ρ__out`

where A is the cross-section area of the valve and `ρ__out` is the fluid density at the exit.


The kinetic energy term in the First Law of Thermodynamics is generally small and is normally be ignored - this makes the combined system of equations explicit, and simple to solve. For this analysis, however, the kinetic energy term will remain. The equations are then implicit, and hence require a numerical solution


restart; with(ThermophysicalData)



fluid := "R717":

Enthalpy at inlet

h__in := Property(massspecificenthalpy, T = T__in, P = P__in, R717)



Density at inlet

rho__in := ThermophysicalData:-Property(density, T = T__in, P = P__in, R717)



Mass flowrate of refrigerant

m := A*v__in*rho__in




P__out := 2*10^5:

Enthalpy and density at outlet

eq1 := h__out = ThermophysicalData:-Property("massspecificenthalpy", "temperature" = T__out, "P" = P__out, fluid)

h__out = ThermophysicalData:-Property("massspecificenthalpy", "temperature" = T__out, "P" = 200000, "R717")


eq2 := rho__out = ThermophysicalData:-Property("D", "temperature" = T__out, "P" = P__out, fluid)

rho__out = ThermophysicalData:-Property("D", "temperature" = T__out, "P" = 200000, "R717")


Mass Conservation and First Law of Thermodynamics

eq3 := h__in+(1/2)*v__in^2 = h__out+(1/2)*v__out^2:

eq4 := m = A*v__out*rho__out:

Solution of the Equation System

res := fsolve({eq1, eq2, eq3, eq4})

{T__out = 285.0179004, h__out = 1654113.289, v__out = 55.27490598, rho__out = 1.474688637}


Plot Thermodynamic Process on a PhT Chart

assign(res); -1; p1 := plot(`~`[`*`](10^(-3), [[h__in, P__in], [h__out, P__out]]), color = black, thickness = 5); -1; p2 := plot(`~`[`*`](10^(-3), [[h__in, P__in], [h__out, P__out]]), thickness = 5, style = point, symbol = solidcircle, color = red, symbolsize = 20); -1; p3 := PHTChart(fluid); -1; plots:-display(p1, p2, p3)